I Boundary and homeomorphism

The 2-sphere ##\mathbb{S}^2## can be expressed as the product ##\mathbb{S}^1 \times \mathbb{S}^1##
Now can we express ##\mathbb{S}^1## as ##\mathbb{S}^1 \subset (-a,a)##, where ##(-a,a)## is some open interval of ##\mathbb{R}##? If so, then (I think) ##\mathbb{S}^1## is homeomorphic to ##\mathbb{R}##. Also, it's homemomorphic to ##\mathbb{R}^2## up to four coordinate charts covering it in ##\mathbb{R}^2##.

If so, by the same reasoning ##\mathbb{S}^2 \subset \mathbb{R}^2## is homeomorphic to ##\mathbb{R}^2##. Also, it's homeomorphic to ##\mathbb{R}^3##, for we can define an embedding from it to ##\mathbb{R}^3##.

Finally, if the above is correct, ##\mathbb{S}^2 \times \mathbb{R}## is homeomorphic to ##\mathbb{R}^3##, though it's not compact.

I'm trying to get this because I'm interested in knowing whether ##\mathbb{S}^2 \times \mathbb{R}## as a manifold has a boundary or not.

I was taught long ago that ##S^1\times S^1## is two dimensional torus not sphere.

Oh, I thought the 2-sphere could be represented as ##\mathbb{S}^1 \times \mathbb{S}^1## because each one of the coordinates of a point ##x \in \mathbb{S}^2##, say ##(\theta, \phi)##, obeys a "circle equation".

Staff: Mentor

Oh, I thought the 2-sphere could be represented as ##\mathbb{S}^1 \times \mathbb{S}^1## because each one of the coordinates of a point ##x \in \mathbb{S}^2##, say ##(\theta, \phi)##, obeys a "circle equation".

This is locally (in a neighborhood of each point) true, but not globally (on the entire sphere), because you will have to leave out one of the poles.

I can't really read what's going on here, but the easiest way to show that ##\mathbb{S}^1## is not homeomorphic to an open interval of ##\mathbb{R}## is to note that ##\mathbb{S}^1## is compact whereas an open interval of ##\mathbb{R}## is not. Since homeomorphisms preserve compactness, we immediately get that the two spaces are not homeomorphic.

I can't really read what's going on here, but the easiest way to show that ##\mathbb{S}^1## is not homeomorphic to an open interval of ##\mathbb{R}## is to note that ##\mathbb{S}^1## is compact whereas an open interval of ##\mathbb{R}## is not. Since homeomorphisms preserve compactness, we immediately get that the two spaces are not homeomorphic.

What if we choose a closed interval instead of a open one. A closed interval is compact.

yes, but it's precisely what I was trying to say in post number 1. Pick a closed interval. There's no way of homeomorphicaly map the entire circle to ##\mathbb{R}^2## using an closed interval (at least in that way I did in post #5). So you have to "remove the end points" of your closed interval. Then now you have an open interval that is homeomorphic to ##\mathbb{R}##.

Staff: Mentor

Then letting the "end points", namely the poles out of consideration is forbidden?

It's not forbidden, it simply makes the difference between local and global. As entire set ##S^1 \times S^1 \ncong S^2## as a torus (on the left) isn't a sphere (on the right), because one has a hole and the other has not. But on a small patch you can always apply coordinates like on earth. You simply cannot find a single chart describing both. As manifolds they all have local flat charts patched to a whole atlas, but this doesn't make them equal.

yes, but it's precisely what I was trying to say in post number 1. Pick a closed interval. There's no way of homeomorphicaly map the entire circle to R2R2\mathbb{R}^2 using an closed interval (at least in that way I did in post #5). So you have to "remove the end points" of your closed interval. Then now you have an open interval that is homeomorphic to RR\mathbb{R}.

Can you point out where I'm wrong on the above ?

I think you answered your own question. Look here:https://www.physicsforums.com/threads/why-the-circle-cant-be-homeomorphic-to-a-real-interval.537897/http://mathhelpforum.com/differenti...-not-homeomorphic-any-interval-real-line.html
Combining with what I said before: an open interval on ##\mathbb{R}## is not homeomorphic to ##\mathbb{S}^1## because compactness isn't preserved. A half-open interval won't work for the same reason. And as you said, there's no homeomorphism between a closed interval of ##\mathbb{R}## and ##\mathbb{S}^1##. The links above answer why: choose a closed real interval ##[a,b]## and assume a (edit: bijection homeomorphism) exists ##f: [a,b] \mapsto \mathbb{S}^1##. Then deleting any point ##p## (not equal to ##a## or ##b##) from the real interval gives another homeomorphism: ##g: [a,p) \cup (p,b] \mapsto \mathbb{S}^1 - \{f(p)\}##. But the new real interval ##[a,p)\cup(p,b]## is not connected, whereas ##\mathbb{S}^1 - f(p)## is connected. Since connectedness is a topological property, ##g## can't be a homeomorphism (and by extension, neither can ##f##).

So can we apply the definition of connectness to intervals like the one above? I thought the definition could be applied only to open intervals. Also, doing the same as you did, but with an open interval ##(a,b)##, the same reasoning would led us to conclude that ##\mathbb{S}^1## and ##(a,b)## are not homeomorphic. Cool!

There is another funny proof that ##[a,b]## is not homeomorphic to ##S^1##. Indeed, any continuous function ##f:[a,b]\to[a,b]## has a fixed point. But there is a continuous function of ##S^1## to itself that does not have a fixed point.

The 2-sphere ##\mathbb{S}^2## can be expressed as the product ##\mathbb{S}^1 \times \mathbb{S}^1##
Now can we express ##\mathbb{S}^1## as ##\mathbb{S}^1 \subset (-a,a)##, where ##(-a,a)## is some open interval of ##\mathbb{R}##? If so, then (I think) ##\mathbb{S}^1## is homeomorphic to ##\mathbb{R}##. Also, it's homemomorphic to ##\mathbb{R}^2## up to four coordinate charts covering it in ##\mathbb{R}^2##.

If so, by the same reasoning ##\mathbb{S}^2 \subset \mathbb{R}^2## is homeomorphic to ##\mathbb{R}^2##. Also, it's homeomorphic to ##\mathbb{R}^3##, for we can define an embedding from it to ##\mathbb{R}^3##.

Finally, if the above is correct, ##\mathbb{S}^2 \times \mathbb{R}## is homeomorphic to ##\mathbb{R}^3##, though it's not compact.

I'm trying to get this because I'm interested in knowing whether ##\mathbb{S}^2 \times \mathbb{R}## as a manifold has a boundary or not.

I'll appreciate any help.

There is a result, I think by Ulam, that ## \mathbb S^n ## cannot be embedded in ## \mathbb R^n ## -or-lower. Still, it would be interesting to see whether ## \mathbb S^n ## " Is a square root" for some n , meaning whether it is homeomorphic to the product ## Y \times Y ## of some topological space ## Y ## . There is an argument showing ## \mathbb R^{2n+1} ## is not a square root in this sense.

There is another funny proof that ##[a,b]## is not homeomorphic to ##S^1##. Indeed, any continuous function ##f:[a,b]\to[a,b]## has a fixed point. But there is a continuous function of ##S^1## to itself that does not have a fixed point.

Nice. And there is also the connectivity one that any point p separates [a,b] , i.e., for any p in [a,b] , [a,b] -{p} is disconnected, while the same is not true for ## \mathbb S^1 ## - or higher-dimensions. I don't know the formal name for this, though ; it comes down to that if X is homeo to Y through h, then X-{p} is homeo to Y-{h(p)}. EDIT 2, besides, with heavier machinery, you also have contractibility, etc. which is preserved under homotopy equivalence alone. EDIT 3:

Staff: Mentor

There is another funny proof that ##[a,b]## is not homeomorphic to ##S^1##. Indeed, any continuous function ##f:[a,b]\to[a,b]## has a fixed point. But there is a continuous function of ##S^1## to itself that does not have a fixed point.

Nice idea. Or one can take ##S^1 \times S^1 \cong \mathbb{P}^2(1,\mathbb{R}) \cong \mathbb{P}(2,\mathbb{R}) \ncong \mathbb{P}(1,\mathbb{C}) \cong S^2\;##, which indicates the problem with the poles.